A population of values has a normal distribution with $$ \mu=145.2 $$ and $$ \sigma=91.6 $$. You intend to draw a random sample of size $$ n=204 $$. Find the probability that a single randomifsselected value is between 133 and 161.9 . $$ P(133

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Let the population be normally distributed with
$$
\mu=145.2 \quad \text{and} \quad \sigma=91.6.
$$

We are given a sample size of
$$
n = 204.
$$

### (a) Single Random Value

We wish to find
$$
P(133 < X < 161.9).
$$

1. **Calculate the $$ z $$-scores for the endpoints.**

For $$ X=133 $$:
   $$
   z_1=\frac{133-\mu}{\sigma} = \frac{133-145.2}{91.6} = \frac{-12.2}{91.6}\approx -0.1333.
   $$

For $$ X=161.9 $$:
   $$
   z_2=\frac{161.9-\mu}{\sigma} = \frac{161.9-145.2}{91.6} = \frac{16.7}{91.6}\approx 0.1823.
   $$

2. **Write the probability in terms of the standard normal CDF, $$\Phi(z)$$:**

$$
   P(133 < X < 161.9)=\Phi(0.1823)-\Phi(-0.1333).
   $$

Since
   $$
   \Phi(-z)=1-\Phi(z),
   $$
   we have
   $$
   \Phi(-0.1333)=1-\Phi(0.1333).
   $$

Thus,
   $$
   P(133 < X < 161.9)=\Phi(0.1823)-\bigl[1-\Phi(0.1333)\bigr] = \Phi(0.1823)+\Phi(0.1333)-1.
   $$

3. **Use approximate values for $$\Phi(z)$$:**

- For $$ z \approx 0.1333 $$, $$\Phi(0.1333)$$ is approximately $$0.5520$$.
   - For $$ z \approx 0.1823 $$, $$\Phi(0.1823)$$ is approximately $$0.5727$$.

Then,
   $$
   P(133 < X < 161.9) \approx 0.5727 + 0.5520 - 1 = 0.1247.
   $$

Rounding to 4 decimal places,
   $$
   P(133 < X < 161.9) \approx 0.1247.
   $$

### (b) Sample Mean

The sample mean $$ M $$ of $$ n $$ independent observations is normally distributed with
$$
M \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right).
$$

1. **Calculate the standard error:**

$$
   \text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{91.6}{\sqrt{204}}.
   $$
   
   First, evaluate
   $$
   \sqrt{204}\approx 14.2829.
   $$
   
   Hence,
   $$
   \text{SE}\approx \frac{91.6}{14.2829}\approx 6.410.
   $$

2. **Find the $$ z $$-scores for the limits for $$ M $$:**

For $$ M=133 $$:
   $$
   z_{\text{low}} = \frac{133-\mu}{\text{SE}} = \frac{133-145.2}{6.410} = \frac{-12.2}{6.410}\approx -1.903.
   $$

For $$ M=161.9 $$:
   $$
   z_{\text{up}} = \frac{161.9-\mu}{\text{SE}} = \frac{161.9-145.2}{6.410} = \frac{16.7}{6.410}\approx 2.605.
   $$

3. **Write the probability:**

$$
   P(133 < M < 161.9) = \Phi(2.605) - \Phi(-1.903).
   $$

4. **Express $$\Phi(-1.903)$$ in terms of a positive $$ z $$-score:**

$$
   \Phi(-1.903) = 1-\Phi(1.903).
   $$

Therefore,
   $$
   P(133 < M < 161.9) = \Phi(2.605) + \Phi(1.903) - 1.
   $$

5. **Use approximate values for $$\Phi(z)$$:**

- For $$ z\approx 2.605 $$, $$\Phi(2.605)$$ is approximately $$0.9954$$.
   - For $$ z\approx 1.903 $$, $$\Phi(1.903)$$ is approximately $$0.9713$$.

Then,
   $$
   P(133 < M < 161.9) \approx 0.9954 + 0.9713 -1 = 0.9667.
   $$

Rounding to 4 decimal places,
   $$
   P(133 < M < 161.9) \approx 0.9667.
   $$

### Final Answers

$$
P(133 < X < 161.9) \approx 0.1247
$$
$$
P(133 < M < 161.9) \approx 0.9667
$$
The probability that a single random value is between 133 and 161.9 is approximately 0.1247. The probability that a sample mean of size 204 is between 133 and 161.9 is approximately 0.9667.

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